Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $n = \dfrac{a + 6}{-3a + 24} \times \dfrac{a^2 - 7a - 8}{2a^2 + 12a} $
Explanation: First factor the quadratic. $n = \dfrac{a + 6}{-3a + 24} \times \dfrac{(a - 8)(a + 1)}{2a^2 + 12a} $ Then factor out any other terms. $n = \dfrac{a + 6}{-3(a - 8)} \times \dfrac{(a - 8)(a + 1)}{2a(a + 6)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (a + 6) \times (a - 8)(a + 1) } { -3(a - 8) \times 2a(a + 6) } $ $n = \dfrac{ (a + 6)(a - 8)(a + 1)}{ -6a(a - 8)(a + 6)} $ Notice that $(a + 6)$ and $(a - 8)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ \cancel{(a + 6)}(a - 8)(a + 1)}{ -6a\cancel{(a - 8)}(a + 6)} $ We are dividing by $a - 8$ , so $a - 8 \neq 0$ Therefore, $a \neq 8$ $n = \dfrac{ \cancel{(a + 6)}\cancel{(a - 8)}(a + 1)}{ -6a\cancel{(a - 8)}\cancel{(a + 6)}} $ We are dividing by $a + 6$ , so $a + 6 \neq 0$ Therefore, $a \neq -6$ $n = \dfrac{a + 1}{-6a} $ $n = \dfrac{-(a + 1)}{6a} ; \space a \neq 8 ; \space a \neq -6 $